📂Odinary Differential Equations

Separable First-Order Differential Equations

Definition¹

A first-order differential equation is said to be separable if it satisfies the following condition:

$$ f(x)+g(y)\dfrac{dy}{dx}=0 \quad \text{or} \quad f(x)dx = -g(y)dy $$

Explanation

It can be expressed in various forms, but the important point is that the variables on each side must be separated. The method of finding solutions by separating these two variables is called the method of separation of variables.

The separability is a very good condition; if a given differential equation is separable, its solution can be easily obtained. On the other hand, if it is not separable, various methods are used to make it into a separable form. In other words, there are various ways to solve a first-order differential equation, but essentially, it all comes down to separation of variables.

Solution

$$ \begin{align*} && g(y)\dfrac{dy}{dx} + f(x)&=0 \\ \implies && g(y)\dfrac{dy}{dx} &=-f(x) \\ \implies && g(y)dy &=-f(x)dx \\ \implies && \int g(y)dy& =-\int f(x)dx+C \end{align*} $$

Here, $C$ is the integration constant. After integration, the left-hand side should be arranged with respect to $y$.

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Example

Find the general solution of $\dfrac{dy}{dx}+y=0$.

$$ \begin{align*} &&\dfrac{dy}{dx}& =-y \\ \implies && \dfrac{1}{y}dy&=-dx \\ \implies && \int \dfrac{1}{y} dy &=-\int dx \\ \implies &&\ln y &=-x+C \\ \implies && y&=e^{-x+C}=e^{-x}e^C=Ce^{-x} \end{align*} $$ If the initial value is $y(0)=y_{0}$, then because of $y(0)=C=y_{0}$, we have

$$ y(x)=y_{0}e^{-x} $$

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Radioactive Decay of Nuclei

The number of radioactive nuclei decaying per unit time is proportional to the number of nuclei $N$.

$$ \dfrac{dN}{dt}=-\lambda N $$

Here, $\lambda$ is the decay constant.

$$ \begin{align*} && \dfrac{dN}{dt} &=-\lambda N \\ \implies && \dfrac{1}{N}dN&=-\lambda dt \\ \implies && \int \dfrac{1}{N}dN &=-\int \lambda dt \\ \implies && \ln N &=-\lambda t+C \\ \implies && N&=Ce^{-\lambda t} \end{align*} $$

If the initial value is $N(0)=N_{0}$, then because of $N(0)=C=N_{0}$, we have

$$ N(t)=N_{0}e^{-\lambda t} $$

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